# Hi everyone! I need help!

1 view (last 30 days)
Engdaw Chane on 9 Apr 2018
Commented: Engdaw Chane on 11 Apr 2018
I need to get the probability of every element in a 3D matrix (Mat) which is 83x92x80
% code
%
[r, c, t] = size(Mat);
y = zeros(r,c,t);
p = zeros(r,c,t);
for i = 1:c;
for j = 1:r;
for k=1:t;
y(j,i,k) = sum(Mat (:,:,[k]) == Mat(j,i,k));
p(j,i,k) = y(j,i,k)/80;
end
end
end
First, I am getting an error “Assignment has more non-singleton rhs dimensions than non-singleton subscripts”. Second, I am not quite sure if this is how I should do what I want to do. I really appreciate your help.
Many thanks
Endaw
Bob Thompson on 9 Apr 2018
The error occurs because your code is trying to fit a three dimensional array, created by the sum() command, into a single element. If you're looking for the total summation, James suggested a decent solution.

James Tursa on 9 Apr 2018
Maybe adding another sum gets the result you want?
y(j,i,k) = sum(sum(Mat (:,:,[k]) == Mat(j,i,k)));
Engdaw Chane on 11 Apr 2018
My matrix has 83 rows, 92 columns and 80 months. What I need is the probability of each value along time (80 months). For example, this is how I calculated the probability of zero values for every row and column along time.
% code
sum(matrix == 0, 3) ./ (size(matrix,3)); % here the result was just a matrix (83x92).
%
Now, I have to calculate the probability of all values.
For example if a value (at a specific row and column) occurred once in all 80 files, the probability for that value would be 1/80=0.0125. But this probability value changes if that specific value occurred multiple times in the 80 files.
thank you for your effort to save me!
Kindly, Chane

Steven Lord on 10 Apr 2018
Are you trying to compute the histogram of the elements in that array? If so take a look at the histogram function (or the histcounts function if you just need the counts without the picture.)
Engdaw Chane on 11 Apr 2018
Steven Lord
This is what I need to do; My matrix has 83 rows, 92 columns and 80 months. What I need is the probability of each value along time (80 months). For example, this is how I calculated the probability of zero values for every row and column along time.
% code
sum(matrix == 0, 3) ./ (size(matrix,3)); % here the result was just a matrix (83x92).
%
Now, I have to calculate the probability of all values.
For example if a value (at a specific row and column) occurred once in all 80 files, the probability for that value would be 1/80=0.0125. But this probability value changes if that specific value occurred multiple times in the 80 files.
thank you for your effort to save me!
Kindly, Chane